= Solution
The two stated formulas show that $Q$ rotates the $(|\beta\rangle,|\alpha\rangle)$ plane by $2\theta$. Starting from $|\xi\rangle$, whose angle from $|\beta\rangle$ is $\theta$, two iterations give angle
$$
\theta+2(2\theta)=5\theta=\frac\pi2.
$$
Hence $Q^2|\xi\rangle=|\alpha\rangle$. The reflection about $|\xi\rangle$ is implementable as
$$
I_\xi=U(I-2|0\cdots0\rangle\langle0\cdots0|)U^\dagger.
$$
\b[Thus apply $U$ to $|0\cdots0\rangle$, followed by two applications of $-I_\xi I_A$. This prepares $|\alpha\rangle$ exactly and deterministically.]
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